AMC 10 · 2013 · #20
Grade 8 geometry-2dA unit square is rotated 45∘ about its center. What is the area of the region swept out by the interior of the square?
Pick an answer.
AMC 10 2013 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: A unit square is turned $45^\circ$ about its center. As it turns, its inside passes over a region of the plane. Find the area of that swept region.
Givens: The shape is a square with side length $1$.; It rotates $45^\circ$ around its own center.; The region wanted is every point the square's interior touches at some moment during the turn.; Answer choices: (A) $1-\frac{\sqrt2}{2}+\frac{\pi}{4}$, (B) $\frac12+\frac{\pi}{4}$, (C) $2-\sqrt2+\frac{\pi}{4}$, (D) $\frac{\sqrt2}{2}+\frac{\pi}{4}$, (E) $1+\frac{\sqrt2}{4}+\frac{\pi}{8}$
Unknowns: The total area covered by the interior of the square across the whole $45^\circ$ rotation.
Understand
Restated: A unit square is turned $45^\circ$ about its center. As it turns, its inside passes over a region of the plane. Find the area of that swept region.
Givens: The shape is a square with side length $1$.; It rotates $45^\circ$ around its own center.; The region wanted is every point the square's interior touches at some moment during the turn.; Answer choices: (A) $1-\frac{\sqrt2}{2}+\frac{\pi}{4}$, (B) $\frac12+\frac{\pi}{4}$, (C) $2-\sqrt2+\frac{\pi}{4}$, (D) $\frac{\sqrt2}{2}+\frac{\pi}{4}$, (E) $1+\frac{\sqrt2}{4}+\frac{\pi}{8}$
Plan
Primary tool: #7 Identify Subproblems
Secondary: #17 Visualize Spatial Relationships, #1 Draw a Diagram, #14 Extreme Principle
The swept blob has a curvy outline and no single formula for it, so Tool #7 (Identify Subproblems) is the engine: cut the region into pieces whose areas we already know how to find. The cut is guided by Tool #17 (Visualize Spatial Relationships) — picturing how each corner traces a circular arc while the sides stay closer in — and Tool #14 (Extreme Principle), which says the farthest any point travels is the corner distance, so that distance is the outer radius. The whole picture is unchanged by a $90^\circ$ turn, and that symmetry lets us solve just two representative pieces (one round, one flat) and multiply.
Execute — Answer: C
8.G.B.7 Step 1 Put it on axes, find the reach
- Place the center $O$ at the origin with the square's corners at $\left(\pm\frac12,\pm\frac12\right)$.
- The distance from the center to a corner is the hypotenuse of a right triangle with legs $\frac12$ and $\frac12$, so it is $\sqrt{\left(\frac12\right)^2+\left(\frac12\right)^2}=\frac{\sqrt2}{2}$, while the center-to-side distance is only $\frac12$.
- When the square turns, each corner — the farthest part — sweeps a circle of radius $\frac{\sqrt2}{2}$, and no point ever gets farther out than that.
💡 Corners are the farthest points from the center, so they carve the outer rim of the sweep.
8.G.A.1 Step 2 Slice into eight wedges
- Draw eight rays out of the center, one every $45^\circ$.
- Turning the finished swept shape a quarter turn ($90^\circ$) leaves it looking exactly the same, so the eight wedges repeat in just two kinds: four “arc wedges” whose outer edge is a corner's circular arc at radius $\frac{\sqrt2}{2}$, and four “flat wedges” whose outer edge is a straight side of the square.
- Find one of each type, then multiply by four.
💡 A shape with $90^\circ$ symmetry has only a couple of distinct pieces — solve those and copy.
7.G.B.4 Step 3 The four arc wedges make a half-disk
- Each arc wedge is a $45^\circ$ pie slice of the circle of radius $\frac{\sqrt2}{2}$ that a corner sweeps.
- Four of them together span $4\times45^\circ=180^\circ$, which is exactly half of that full circle.
- So the four arc wedges combine into a semicircle of radius $\frac{\sqrt2}{2}$, with area $\frac12\pi\left(\frac{\sqrt2}{2}\right)^2=\frac12\pi\cdot\frac12=\frac{\pi}{4}$.
💡 Four equal $45^\circ$ pie slices of the same radius glue into a half pie.
6.G.A.3 Step 4 One flat wedge by coordinates
- Take the flat wedge in the first octant.
- Its four corners are the center $O=(0,0)$; the turned square's vertex $P_1=\left(\frac{\sqrt2}{2},0\right)$; the point $Q=\left(\frac12,\ \frac{\sqrt2}{2}-\frac12\right)$ where the starting square's side $x=\frac12$ crosses the turned square's side $x+y=\frac{\sqrt2}{2}$; and the starting square's vertex $P_2=\left(\frac12,\frac12\right)$.
- The shoelace formula on $O,P_1,Q,P_2$ gives area $\frac{2-\sqrt2}{4}$, so the four flat wedges total $4\cdot\frac{2-\sqrt2}{4}=2-\sqrt2$.
💡 A wedge with straight sides is just a polygon — its corners' coordinates fix its area.
6.G.A.1 Step 5 Add the round part and the flat part
- The swept region is the four arc wedges plus the four flat wedges.
- Adding the round part $\frac{\pi}{4}$ to the flat part $2-\sqrt2$ gives $2-\sqrt2+\frac{\pi}{4}$.
- Numerically that is about $2-1.414+0.785\approx1.37$.
- This matches choice $\textbf{(C)}$.
💡 Total area is the round pieces plus the straight pieces, nothing double-counted.
8.G.B.7 Place the center $O$ at the origin with the square's corners at $\left(\pm\frac1 8.G.A.1 Draw eight rays out of the center, one every $45^\circ$. Turning the finished sw 7.G.B.4 Each arc wedge is a $45^\circ$ pie slice of the circle of radius $\frac{\sqrt2}{ 6.G.A.3 Take the flat wedge in the first octant. Its four corners are the center $O=(0,0 6.G.A.1 The swept region is the four arc wedges plus the four flat wedges. Adding the ro Review
Reasonableness: The answer must beat the square's own area, since the starting square ($0^\circ$) is part of the sweep: indeed $1.37>1$. It must also stay below the disk of radius $\frac{\sqrt2}{2}$ that contains everything, whose area is $\pi\left(\frac{\sqrt2}{2}\right)^2=\frac{\pi}{2}\approx1.57$: indeed $1.37<1.57$. So $1<2-\sqrt2+\frac{\pi}{4}<\frac{\pi}{2}$, exactly where a $45^\circ$ sweep should land. The nearby distractors miss: (B) $\approx1.29$ is too small (its flat part $\frac12$ undercounts $2-\sqrt2\approx0.59$) and (D) $\approx1.49$ is too big, so only (C) fits both bounds.
Alternative: Count from the square outward instead. The swept region is the whole unit square plus the four bulges each corner pushes beyond a side; those bulges together add $1-\sqrt2+\frac{\pi}{4}$, and $1+\left(1-\sqrt2+\frac{\pi}{4}\right)=2-\sqrt2+\frac{\pi}{4}$, the same result. Equivalently, integrate the outer radius: $\frac12\int_0^{2\pi}R(\varphi)^2\,d\varphi$ with $R$ periodic every $90^\circ$ splits into the same $\frac{\pi}{4}$ (where $R=\frac{\sqrt2}{2}$) plus $2-\sqrt2$ (where the sides bound it).
CCSS standards used (min grade 8)
8.G.B.7Apply the Pythagorean theorem to determine unknown side lengths in right triangles (Finding the center-to-corner distance $\frac{\sqrt2}{2}$ (the outer sweep radius) from legs $\frac12$ and $\frac12$.)8.G.A.1Verify experimentally the properties of rotations, reflections, and translations (Seeing that each corner traces a circular arc under rotation and that the swept figure is unchanged by a $90^\circ$ turn, so its wedges repeat in four congruent copies.)7.G.B.4Know the formulas for area and circumference of a circle (Computing the four arc wedges as a semicircle of radius $\frac{\sqrt2}{2}$, area $\frac{\pi}{4}$.)6.G.A.3Draw polygons in the coordinate plane given coordinates for the vertices (Placing the flat wedge's vertices $O,P_1,Q,P_2$ on axes and applying the shoelace formula for its area.)6.G.A.1Find area of triangles, special quadrilaterals, and polygons by composing (Adding the round part $\frac{\pi}{4}$ and the flat part $2-\sqrt2$ into the total swept area.)
⭐ Spin the square and split the trail into eight wedges: four rounded ones fold into a half-circle of radius $\frac{\sqrt2}{2}$ ($\frac{\pi}{4}$) and four straight ones add up to $2-\sqrt2$, for a total of $2-\sqrt2+\frac{\pi}{4}$, choice $\textbf{(C)}$.
⭐ Spin the square and split the trail into eight wedges: four rounded ones fold into a half-circle of radius $\frac{\sqrt2}{2}$ ($\frac{\pi}{4}$) and four straight ones add up to $2-\sqrt2$, for a total of $2-\sqrt2+\frac{\pi}{4}$, choice $\textbf{(C)}$.
More like this
Same archetype — closest grade level first.
